Question

The ratio of number of moles of $KMn{O}_4$ and $K_2{Cr}_2{O}_7$ required to oxidize 0.1 mol ${Sn}^{2+}$ to ${Sn}^{+4}$ in acidic medium:

• A. 6:5
• B. 5:6
• C. 1:2
• D. 2:1

Answer 1

Answer: (A)

6:5

2 $KMn{O}_4$ + 5 $S{n}^{2+}$ + 16 $H^{+}\to$ 5 $S{n}^{4+}$ + 2 $M{n}^{2+}$ + 8$H_{2}O$ + 2 $K^{+}$

(0.1 mol $S{n}^{2+}$) $\times$ (2 $KMn{O}_4$ / 5 $S{n}^{2+}$)= 0.04 mol $KMn{O}_4$

=$K_2{Cr}_2{O}_7$ + 3 $S{n}^{2+}$ + 14 $H^{+}$ $\to$ 3 $S{n}^{4+}$ + 2 $C{r}^{3+}$ + 7$H_{2}O$ + 2 $K^{+}$

(0.1 mol $S{n}^{2+}$) $\times$ (1$K_2{Cr}_2{O}_7$ / 3 $S{n}^{2+}$)= 0.03333 mol $K_2{Cr}_2{O}_7$

(0.04 mol $KMn{O}_4$ / 0.03333 mol $K_2{Cr}_2{O}_7)0.12/0.1$

= 1.2

= 6/5

6 : 5